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Differential Equations

JEE Main 2025 (Online) 28th January Evening Shift

INTEGER+4 / -12025

If $y=y(x)$ is the solution of the differential equation,
$\sqrt{4-x^2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^2-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^2-8}{4}$, then $y^2(0)$ is equal to ___________.

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